Problem: Find $\lim_{x\to 0}(2e^x-1)^{^{{\frac{1}{x^2+2x}}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $e$ (Choice C) C $\dfrac e4$ (Choice D) D The limit doesn't exist.
Answer: Substituting $x=0$ into $(2e^x-1)^{^{{\frac{1}{x^2+2x}}}}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(2e^x-1)^{^{{\frac{1}{x^2+2x}}}}$, we will find $\lim_{x\to 0}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 0}y$. $\ln(y) =\dfrac{\ln(2e^x-1)}{x^2+2x}$ Substituting $x=0$ into $\dfrac{\ln(2e^x-1)}{x^2+2x}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 0}\ln(y) \\\\ &=\lim_{x\to 0}\dfrac{\ln(2e^x-1)}{x^2+2x} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(2e^x-1)]}{\dfrac{d}{dx}[x^2+2x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{\left(\dfrac{2e^x}{2e^x-1}\right)}{2x+2} \\\\ &=\dfrac{\left(\dfrac{2}{1}\right)}{2} \gray{\text{Substitution}} \\\\ &=1 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(2e^x-1)]}{\dfrac{d}{dx}[2x+2]}$ actually exists. We found that $\lim_{x\to 0}\ln(y)=1$, which means $\lim_{x\to 0}y=e$. [Why?] In conclusion, $\lim_{x\to 0}(2e^x-1)^{^{{\frac{1}{x^2+2x}}}}=e$.